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Let $f:X\longrightarrow Y$ be a map between CW-complexes $X$ and $Y$. By the Whitehead Theorems, if one of the conditions:

1- (homotopy version) $\pi_n (f):\pi_n (X)\longrightarrow \pi_n (Y)$ is an isomorphism for all $n\geq 1$,

or

2- (homology version) $\pi_1 (f):\pi_1 (X)\longrightarrow \pi_1 (Y)$ and $H_n (\tilde{f}):H_n (\tilde{X})\longrightarrow H_n (\tilde{Y})$ are isomorphisms for all $n\geq 2$,

hold, then there is a map $g:Y\longrightarrow X$ such that $g\circ f\simeq id_X$ and $f\circ g\simeq id_Y$.

Question: Is there any weaker condition (with respect to above conditions) under which there is a map $g:Y\longrightarrow X$ such that we have only $g\circ f\simeq id_X$?

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  • 1
    $\begingroup$ This would not be a weak version of the Whitehead theorem in that it doesn’t follow from the Whitehead theorem. Actually I think it will be much harder to say anything about this. $\endgroup$ – Qiaochu Yuan Nov 18 '17 at 8:41
  • $\begingroup$ You probably want to mumble connected somewhere $\endgroup$ – Thomas Rot Nov 18 '17 at 9:01
  • $\begingroup$ @QiaochuYuan Thank you for the comment. I agree with you. $\endgroup$ – M.Ramana Nov 18 '17 at 9:03
  • $\begingroup$ @ThomasRot I don't understand your mean. $\endgroup$ – M.Ramana Nov 18 '17 at 9:12
  • $\begingroup$ (2) is true if $X$ and $Y$ are simply connected, or more generally simple, or even more generally nilpotent spaces, but not for general CW complexes. (On the other hand, for nilpotent spaces you do not need to require $\pi_1$-isomorphism, it follows form the homology isomorphism). $\endgroup$ – Gregory Arone Nov 18 '17 at 11:06
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Suppose you have a map $f\colon X\to Y$ of finite CW complexes such that $K(p,n)_*(f)$ is injective for all primes $p$ and integers $n\geq 0$ (where $K(p,n)$ is Morava $K$-theory). Then the Nilpotence Theorem of Hopkins, Devinatz and Smith implies that the map $\Sigma^kf^{(m)}\colon \Sigma^kX^{(m)}\to\Sigma^kY^{(m)}$ has a left inverse $g$ for sufficiently large $k$ and $m$. This may not seem very satisfactory but I think that it is the best that you can reasonably hope to do. The question of whether a map has a left inverse is just intrinsically much more subtle than the question of whether it is an equivalence.

One can also say that in the category of finite spectra, a map $f\colon X\to Y$ has a left inverse iff $\pi_*^S(f)\colon \pi_*^S(X)\to\pi_*^S(Y)$ is injective, provided that Freyd's Generating Hypothesis is correct. But the Generating Hypothesis has been any open question for fifty years now, which is another indication that the problem is intrinsically hard.

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  • $\begingroup$ Thank you very much for your answer. This can be so useful for working on these problems. Thanks for more explanations. $\endgroup$ – M.Ramana Nov 19 '17 at 4:20

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